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Oduvanchick [21]
3 years ago
12

A greyhound's velocity changes from 10 meters per second to 15 meters per second in 0.5 second. What is the greyhound's average

acceleration? 5 m/s2 10 m/s 10 m/s2 15 m/s
Physics
1 answer:
Elden [556K]3 years ago
4 0

The answer would be 10ms 2

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The propeller of an airplane is at rest when the pilot starts the engine; and its angular acceleration is a constant value. Two
Maurinko [17]

Answer:0.318 revolutions

Explanation:

Given

Initially Propeller is at rest i.e. \omega _0=0 rad/s

after t=10 s

\omega =10 rad/s

using \omega =\omega _0+\alpha t

10=0+\alpha \cdot 10

\alpha =1 rad/s^2

Revolutions turned in 2 s

\theta =\omega _0t+\frac{\alpha t^2}{2}

\theta =0+\frac{1\times 2^2}{2}

\theta =2 rad

To get revolution \frac{\theta }{2\pi }

=\frac{2}{2\pi}=0.318\ revolutions

3 0
3 years ago
Read 2 more answers
The normal boiling point of a certain liquid is , but when of urea () are dissolved in of the solution boils at instead. Use thi
Kruka [31]

Answer:

100 Degrees is boiling point.

Explanation:

5 0
2 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

5 0
3 years ago
A machine can never be 100% efficient because some work is always lost due to .
klasskru [66]
Friction
Hope it helped
5 0
3 years ago
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a body is moving with uniform acceleration, has initial velocity 45km/hr. and acceleration 20cm/s^2. find its velocity after 25
Eddi Din [679]
That’s hard wow!!!!!!
7 0
3 years ago
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